Mathematics - Hyperbola Question with Solution | TestHub

Given hyperbola is
$\begin{array}{rl}& 9x^2-16y^2-36x+96y-252=0\\ & \Rightarrow 9\left(x^2-4x\right)-16\left(y^2-6y\right)=252\\ & \Rightarrow 9\left(x^2-4x+4\right)-16\left(y^2-6y+9\right)\\ & =252+36-144\\ & \Rightarrow 9(x-2{)}^2-16(y-3{)}^2=144\end{array}$


Put $x=X+2$ and $y=Y+3$
$\therefore$ Equation (i) becomes $\frac{X^2}{16}-\frac{Y^2}{9}=1$
Now, vertices are $X=\pm a$ wehre $a=4$ and $Y=0$
Hence, vertices are $(6,3),(-2,3)$.